---恢复内容开始---
Given the value of a+b and ab you will have to find the value of an+bn
给出a+b和a*b的值，再给出n求a^n+b^n的值.
Input
The input file contains several lines of inputs. Each line except the last line contains 3 non-negative integers p, q and n. Here p denotes the value of a+b andq denotes the value of ab. Input is terminated by a line containing only two zeroes. This line should not be processed. Each number in the input file fits in a signed 32-bit integer. There will be no such input so that you have to find the value of 00.
多组测试数据，每组给出3个数，依次为p、q、n。p代表a+b，q代表a*b。当p和q都为0是这组数组不处理。
Output
For each line of input except the last one produce one line of output. This line contains the value of an+bn. You can always assume that an+bn fits in a signed 64-bit integer.
分析：
定义f(n)=an+bn,则有f(n)∗(a+b)=(an+bn)∗(a+b)=an+1+abn+ban+bn+1=f(n+1)+abf(n−1), 所以f(n+1)=(a+b)f(n)−abf(n−1)
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxsize = 100;
typedef long long ll;
struct matrix{
ll f[2][2];
};
matrix mul(matrix a,matrix b)
{
ll i,j,k;
matrix c;
memset(c.f,0,sizeof(c.f));
for(i=0;i<2;i++)
for(j=0;j<2;j++)
for(k=0;k<2;k++)
c.f[i][j]+=a.f[i][k]*b.f[k][j];
return c;
} matrix ksm(matrix e,ll n)
{
matrix s;
s.f[0][0]=s.f[1][1]=1;
s.f[1][0]=s.f[0][1]=0;
while(n)
{
if(n&1)
s=mul(s,e);
e=mul(e,e);
n=n>>1;
}
return s;
}
matrix e;
int main()
{
ll p,q,n;
while(cin>>p>>q>>n)
{
if(n==0)
{
cout<<2<<endl;
continue;
}
e.f[0][0]=p;
e.f[0][1]=1;
e.f[1][0]=-q;
e.f[1][1]=0;
e=ksm(e,n-1);
ll ans;
ans=p*e.f[0][0]+2*e.f[1][0];
cout<<ans<<endl; }
return 0;
}